I find these results in the evaluation of the logarithms that only differ in the sign $-$ I do not understand why in the first case $\operatorname{Log}[x+1]/8$ is not returned as an answer.
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$\begingroup$ You miss the $+C$ in both $\endgroup$– Claude LeiboviciCommented Jun 16 at 15:24
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$\begingroup$ Actually, that is the format you normally have in Mathematica integrals answers. I did not write the answers. $\endgroup$– Emerson VillafuerteCommented Jun 16 at 15:31
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1 Answer
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This can be understood from logarithm properties. When you take an indefinite integral, there is an arbitrary additive constant at the end, so really we have
$$\int\frac{1}{8(1+x)}dx = \frac{1}{8}\ln(1+x) + C$$
Note that $\ln(8(1+x)) = \ln(8) + \ln(1+x)$, and therefore $\ln(8(1+x))+C_1 = \ln(1+x) + C_2$. So Mathematica's answers are correct: it is just a matter of simplifying it a bit.
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1$\begingroup$ I will add the absolute value in the $\ln$. $\endgroup$ Commented Jun 16 at 15:26